Question

Find an explicit solution of the given initial-value problem.dy/dx = ye^x^2, y(3) = 1.

Answer 1

Answer:

$y=exp(\int\limits^x_4 {e^{-t^{2} } } \, dt)$

Step-by-step explanation:

This is a separable equation with an initial value i.e. y(3)=1.

Take y from right hand side and divide to left hand side ;Take dx from left hand side and multiply to right hand side:

$\frac{dy}{y} =e^{-x^{2} }dx$

Take t as a dummy variable, integrate both sides with respect to "t" and substituting x=t (e.g. dx=dt):

$\int\limits^x_3 {\frac{1}{y} } \, \frac{dy}{dt} dt=\int\limits^x_3 {e^{-t^{2} } } dt$

Integrate on both sides:

$ln(y(t))\left \{ {{t=x} \atop {t=3}} \right. =\int\limits^x_3 {e^{-t^{2} } } \, dt$

Use initial condition i.e. y(3) = 1:

$ln(y(x))-(ln1)=\int\limits^x_3 {e^{-t^{2} } } \, dt\\ln(y(x))=\int\limits^x_3 {e^{-t^{2} } } \, dt$

Taking exponents on both sides to remove "ln":

$y=exp (\int\limits^x_3 {e^{-t^{2} } } \, dt)$